January, 2008
Gains from Manipulating Social Choice Rules
Donald E. Campbell1 and Jerry S. Kelly2
1
Department of Economics and The Program in Public Policy, The College of William and Mary,
Williamsburg, VA 23187-8795
(e-mail: [email protected])
2
Department of Economics, Syracuse University, Syracuse, NY 13244-1020
(e-mail: [email protected])
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Abstract
Gibbard-Satterthwaite analysis is extended by examining the size of the
potential gains that can be achieved from manipulation.
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Gains from Manipulating Social Choice Rules
Donald E. Campbell and Jerry S. Kelly1
Section 1. Introduction.
Section 2. Definitions and first examples.
Section 3. General remarks.
Section 4. Rules with ñ(f) = m but small gain.
Section 5. Condorcet.
Section 6. Scoring rules.
Subsection 6.1. Standard rules with two values.
Subsection 6.2. Standard rules with three values.
Subsection 6.3. Standard rules with m values.
Section 7. Tops-only and vector rules.
Section 8. Final remarks.
Section 1. Introduction.
The Gibbard-Satterthwaite theorem [Gibbard (1973), Satterthwaite (1975)]is a standard
point of entry to the study of implementation theory and mechanism design [e.g., Sonnenschein
(1998), Mas-Colell, Whinston and Green (1995)] and is a core topic in social choice theory. The
theorem tells us that essentially all rules are manipulable, but does not reveal the potential gains
that can be achieved from manipulation or specify consequent losses by others. The present paper
extends the Gibbard-Satterthwaite analysis in the study of potential gains while a companion
paper [Campbell and Kelly (2007)] treats the losses borne by other individuals when one
manipulates for gain.
1
The authors are grateful for an invitation from the University of Athens to visit
their PhD program where some of this work was done.
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This paper is part of a large literature on “degree of manipulability.” A rule may be
thought of as less manipulable if it is manipulable at fewer profiles [Peleg (1979); Nitzan (1985);
Chamberlain (1985); Lepelley and Mbih (1987); Saari (1990); Kelly (1993); Kim and Roush
(1996)].
As a special case, one could seek rules manipulable at a minimum number of profiles
[Kelly (1988); Fristrup and Keiding (1998); Maus, Peters, and Storcken (2005a, 2005b, 2005c,
2005d)]. Alternatively, a rule is less manipulable if there is a smaller number of manipulating
strategies [Chamberlain (1985); Smith (1999); Aleskerov and Kurbanov (1999)], if more
information is required to determine a manipulating strategy [Nurmi (1987)], if the average size of
a manipulating coalition must be larger [Chamberlain (1985); Lepelley and Mbih (1994);
Pritchard and Slinko (2006)], or if there exist countercoalitions to manipulations [Pattanaik
(1976a, 1976b, 1978); Favardin, Lepelley, and Serias (2002)].
Here we explore a different approach: If a rule allows only small gains from manipulation,
we may be persuaded that the cost to the individual of gathering enough information about the
preferences of others to ensure that manipulation is advantageous will not yield a gain once those
costs are factored in. A referee has directed our attention to two papers that have looked at the
possibility of allowing small gains from manipulation, Schummer (2004) and Ehlers, Peters, and
Storcken (2004). Both of these papers work in a very different context than we consider here.
For them, the feasible set is a continuum, ú2+ for Schummer and úm for Ehlers et al. They thus
can not measure gains by number of ranks risen and have to introduce very restricted utility
functions to measure gains: linear for Schummer and Lipschitz continuous for Ehlers et al.
Our analysis assumes a finite number of alternatives, (unrestricted) strong preferences, and
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measures gains by how many ranks one rise up in one’s preference ordering. We are generalizing
Gibbard and Satterthwaite, which says only gain $1, by using only their assumptions (nondictatorship, full domain, and sufficiently large range) while establishing lower bounds on the
possible gains from manipulation greater than 1.
Section 2. Definitions and first examples.
X = {x1, x2, ..., xm} is the finite set of m alternatives. N = {1, 2, ..., n} is the finite set of n
individuals. For any subset Y of X, let ‹(Y) represent the set of all strong (transitive and
asymmetric) orderings on Y. A profile is a function from N into ‹(X) and Ž denotes the set of all
profiles. We let p(h) denote the ordering assigned to person h by profile p. Two profiles are hvariants, where h 0 N, if q(i) = p(i) for all i … h. For strong ordering p(h), we let p(h)[k] denote
the alternative in the kth rank of p(h) and we let p(h)[1:k]denote an initial segment of p(h), the
set of alternatives in the top k ranks of p(h).
A social choice rule f is a function from Ž into X . Note that a full domain assumption is
built in. For a rule f and non-negative integer t, we say f allows a gain of t if there exist two
profiles p and q and an individual h such that:
(1) p and q are h-variants;
(2) f(q) ranks t positions higher in p(h) than f(p).
A rule f has a gain of k, written G(f) = k, if k is the maximum t such that f allows a gain of t.
G(f) is bounded above by m – 1; with no restrictions on the rule f, the gain G(f) can be as
large as m – 1.
Example 2.1. For the rule f that always selects the alternative lowest in individual #1's
ranking, p(1), it is easy to confirm that G(f) = m – 1.
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Just as the Gibbard-Satterthwaite theorem has to distinguish dictatorial from nondictatorial rules, so we will find the gain of a rule to be closely related to the extent to which a
rule favors some individual, by which we mean, in this paper, the extent to which some individual
can precipitate the chosen alternative to be from some initial segment of their ordering. Two
more illustrative examples follow:
Example 2.2. For even n, fix k > 1 and define rule f as follows: For arbitrary p 0 Ž, let Yp
= {p(1)[1], p(1)[k]}. Then f(p) is the alternative in Yp that a majority of the individuals in N\{1}
prefer to the other member of Yp.
Consider profiles p:
1
i>1
x1
x2
xk
x1
x2
!
xk
!
!
1
i>1
x1
xk
xk
x1
x2
and q:
!
x2
!
!
where q differs from p only in that x2 and xk switch places in p(1) to create q(1). Then f(p) = xk
while f(q) = x1. Therefore, G(f) $ k – 1. But G(f) # k – 1 since no i > 1 finds it beneficial to
misrepresent since those individuals are simply choosing from a two-element set by majority rule,
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and so we have G(f) = k – 1. In summary, f(p) is forced to be in p(1)[1:k] and G(f) = k – 1.
Example 2.3. X = {w,x,y,z} and n = 3. At any profile u, at least one alternative is not
anyone’s last. Take f(u) to be the alphabetically2 earliest alternative that no one ranks last3. Then
f(u) can not be u(j)[4] for anyone; so no one can gain more than two ranks. The largest gain an
individual can realize is to go from 3rd to 1st; for this rule, G(f) = 2.
Section 3. General remarks.
The examples of the previous section suggest relating gain to the extent to which
individuals can precipitate the chosen alternative to be from initial segments of their ordering.
For a rule f, let ñi(f) be the smallest integer s such that for all p 0 Ž, f(p) must be in p(i)[1:s], the
initial segment of p(i). A manipulation by i can gain at most ñi(f) – 1 ranks. Since the individual
who manipulates may be an individual with small ñi(f), we set ñ(f) to be the minimum4 over all
i 0 N of the values ñi(f). We will let ñ(f) be a benchmark against which we compare G(f). For
2
If as here, the alternatives are letters of the alphabet, the “alphabetically earliest” is
interpreted in the usual way; when X = {x1, x2, ... , xm} then the “alphabetical” ordering is
x1 < x2 < ... < xm.
3
Example 2.3 is a special case of inverse plurality, axiomatized by Baharad and Nitzan
(2005).
4
For all but a few cases with small n, like Example 2.3, most individuals i will have ñi(f) =
m. For suppose min ñi(f) = k < m and, without loss of generality, let ñ1(f) = k. Then
For example, imagine n = 10, m = 5, ñ1(f) = 3. If ñi(f) < m for as many as three individuals
i > 1, look at a profile p where p(1)[1:3] = {x,y,z}, but each of these alternatives is at the bottom
of someone else’s ordering; an inconsistency. Hence at least seven individuals have ñi(f) = m.
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almost all rules, G(f) $ ñ(f) – 2, but we will see several kinds of rules where G(f) is about half of
ñ(f).
Our analysis begins by dividing into three cases based on the size of ñ(f):
Case 1. ñ(f) = 1. This case is easily disposed of: someone is a dictator and hence
G(f) = 0 ( = ñ(f) – 1).
Case 2. 1 < ñ(f) < m. For this case all our examples will be seen to satisfy the conjecture:
Gain Conjecture #1: For any rule f with |Range(f)| $ 3: if 1 < ñ(f) < m, then
G(f) $ ñ(f) – 1.
A special case of this Conjecture is easy: Suppose that ñ(f) = 2 < m. Since ñ(f) > 1, f is
not dictatorial and the Gibbard-Satterthwaite Theorem tells us G(f) $ 1 = ñ(f) – 1.
It is possible that G(f) is much larger than ñ(f) – 1. Because someone other than the
individual with smallest ñi(f), which determines ñ(f), may be able to manipulate more than that
minimum, ñ(f) – 1.
Example 3.2. For even n, fix k > 1 and define rule f as follows: f(u) is the element within
u(1)[1:s] that is lowest in u(2). ñ1(f) = s while ñi(f) = m for all i > 1. G(f) = m – 1 as individual
#2 can manipulate from his last to his first.
Case 3. ñ(f) = m. This last case is more complicated. G(f) is either m – 1 or m – 2 for
almost all rules with ñ(f) = m. But we will nevertheless see cases where G(f) can be as small as
m/2 – 1.
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Section 4. Rules with ñ(f) = m but small gain.
We illustrate the possibility of small gain with three examples. The first has a range that is
a proper subset of X.
Example 4.1. X = {x1, x2, ..., xm} and let k be an integer between 1 and m – 1 inclusive.
Let f(p) be the alphabetically earliest alternative xj 0 X such that xj is one of the top k alternatives
in the ordering p(i), for some individual i. This is an anonymous rule for which ñ(f) = m. (For
example, if x1 is at the bottom of p(1) and at the top of p(2) then f(p) = x1.) However, we will see
that, depending on the size of k, G(f) can be significantly less than m – 1. Consider profile p:
1
i>1
xm!k+1
xm!k+2
!
xm!1
xm
x2
x1
x3
x4
!
xm!k
x2
!
k
!
m–k
x1
Then f(p) = x2 but person 1 can get x1, which ranks in position k+1 in p(1), by reporting an
ordering with x1 on top. Therefore, G(f) $ m ! k ! 1. We can also see that G(f) $ k ! 1 by
considering profile q for which q(1)[k] = x1:
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1
i>1
x2
!
x1
x2
!
x3
!
!
x1
k
m–k
Here, f(q) = x1; but if individual #1 interchanges x1 and x3, the outcome will be x2 and #1
will have gained k – 1 ranks. In fact, for this f, G(f) = Max [m ! k ! 1, k ! 1]. To see that, it
suffices to observe that no individual has the ability to change the outcome of the rule from one of
his bottom m – k to one of his top k.
Note that there are k – 1 alternatives that are never selected by f. For arbitrary profile p,
the top k alternatives in p(1) must include at least one alternative xj such that j < m–k+2, because
{m–k+2, m–k+3, ... m–1, m} is a set with only k – 1 members. Hence, none of the alternatives
xm–k+2, xm–k+3, ... xm–1, xm belongs to the range of f.
Since either m ! k or k is at least m/2, this example is consistent with the following
conjecture:
Gain Conjecture #2: For any rule f with |Range(f)| $ 3: if ñ(f) = m, then G(f) $ m/2 – 1.
Example 4.2.5 Suppose that n $ 3. Let t1 and t2 be positive integers such that t1 + t2 =
m. At profile u, for each individual i, assign 0 to the t1 alternatives ranked highest in u(i) and 1 to
5
This example is a special case of a standard scoring rule. Scoring rules are treated in
Section 6.
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the t2 alternatives ranked lowest. Then each alternative x in X is assigned the sum of its scores
over all n of the u(i) orderings. f(u) is then the alternative with the smallest sum. Of course, ties
are possible, so a complete description of f requires a tie-break rule. Here we assume that X is
ordered (“alphabetically”) and then f(u) is the alphabetically earliest of the alternatives with the
smallest sum.
Consider now cases where nt2 $ m. Then a profile u can be constructed in which every
alternative is assigned 1 by at least one individual. Let i be an individual for whom f(u) ó
p(i)[1:t1]. Construct new profile u* by bringing f(u) to i’s bottom and having u otherwise
unchanged. f(u*) = f(u) = u*(i)[m]. By anonymity, ñ(f) = m.
If one of individual j’s bottom t2 alternatives is selected under truthful revelation, then j
can not precipitate the selection of one of his top t1 alternatives by misrepresentation because
there is nothing j can do unilaterally to reduce the score of one of those t1 alternatives. So the
possible gains are bounded by the larger of t1–1 or t2–1. We show G(f) = Max [t1–1, t 2–1].
To show a gain of t1 – 1 is possible, consider profile u:
1
2
þ
n
b
x
!
a
b
x
!
a
b
x
!
a
b
x
!
a
z
!
z
!
z
!
z
!
t1
t2
where a is in everyone’s rank t1. Then f(u) = a. But if u* is created by having #1 interchange a
and z, f(u*) = b and #1 gains t1 – 1 positions.
So we turn our attention to the case where t2 – 1 > t1 – 1. We construct a profile u as
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follows:
1
2
þ
n
z
!
!
p
a
b
!
k
a
b
!
k
a
b
!
k
b
!
c
a
l
!
!
z
l
!
!
z
l
!
!
z
t1
t2
where
•
Individuals 2, ..., n all have the same ordering: ab ... k l... z where k appears in
position t1;
•
Individual #1's preferences are a modification of the reverse order z ... lk ... ba.
The modification raise b to rank t1 + 1.
Since t2 > t1, alternatives a, b, ..., k are assigned 1 by individual #1, and those alternatives
get a score of just 1 while the score of the other alternatives is at least n–1 $ 1. By the tie-break,
f(u) = a = u(1)[m]. If u* is obtained from u by interchanging p and b for individual 1, then the
score of b decreases to 0 and f(u*) = b, a gain of t2 – 1 ranks.
The same results could all be achieved if instead of an alphabetic tie-break we had used,
say, individual #2's ranking as the tie-break.
Again, G(f) = Max [t1–1, t2–1] is consistent with Gain Conjecture #2.
Example 4.3. m = 100 and n = 3. Start with a Rawlsian rule: At profile u, for each
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alternative x, determine t(x) = maxi {k : u(i)[k] = x}, the lowest rank occupied by x over the n
individuals. Then f(u) is the alphabetically earliest alternative x with minimum t(x). For m = 100
and n = 3, the largest possible minimum value of t(x) is 67 and hence ñ(f) = 67. Gain Conjecture
#1 would lead us to expect that G(f) $ 66. Consider profile u:
1
2
3
x2
!
x1
x4
!
x2
x5
!
x2
67
x3
x5
!
x98
x3
x6
!
x99
x4
x7
!
x100
33
Alternatives xi for i > 2 have t(xi) > 67, while t(x1) = t(x2) = 67. By the alphabetic tie-break, f(u) =
x1. If u* is obtained by having individual #1 switch x1 and x3, t(x2) = 67 is the minimum and f(u*)
= x2. Thus #1 gains 66 ranks. Since ñ(f) = 67, this 66 is the largest possible gain: G(f) = 66.
Now construct a new rule g, as follows: If at u there is an alternative x that is at the top of
two or three individual rankings, then g(u) = x; otherwise g(u) = f(u). Now ñ(g) = 100 because
one individual’s bottom will be selected if it is at the top for the other two. But, G(g) = 66. To
see this, it suffices to see that no one manipulates away from an alternative that is selected when it
is at two or more tops. Obviously, G(g) = 66 is consistent with Gain Conjecture #2.
Section 5. Condorcet.
In this section we compute the gain to manipulation when we impose the condition that f is
a Condorcet rule. Say x is a strong Condorcet winner if |{i : x ™ y}| > |{i : y ™ x}| for all y in
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X\{x}, that is, x defeats every other alternative under majority voting. Rule f is a Condorcet rule if
for every alternative x and every profile p, if x is a strong Condorcet winner at p, then f(p) = x.
Condorcet rules are non-dictatorial, have range X, and (if n > 2) have ñ(f) = m. Note that for
n = 2, Condorcet does not imply ñ(f) = m; the case n = 2 will be taken up later in this section.
Theorem 5.1 (First Condorcet Gain Theorem). If m $ 3, and n $ 3 is odd, then G(f) $
m – 2 for any Condorcet rule f.
Proof: Suppose that X = {x, y, z, w, . . . }.
Part 1. Assume first that n = 3. Let p be any profile with the three alternatives x, y, and z
arranged at the top in a Latin square, thus:
1
2
3
x
y
z
y
z
x
z
x
y
!
!
!
If f(p) = x then let pN be any profile for which pN(2) = p(2) and pN(3) = p(3), with x ranked
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first in pN(1), y ranked last in pN(1), and z second last in pN(1):
1
2
3
x
y
z
!
z
x
z
x
y
y
!
!
Notice that at p (and pN) both persons 2 and 3 prefer z to every other alternative except y, and that
at pN both persons 1 and 3 prefer z to y. Therefore, z is a Condorcet winner at pN, and hence f(pN)
= z. But z ranks second last in pN(1), x ranks first, and at profile pN person 1 can precipitate the
selection of x by reporting p(1). Therefore, G(f) $ m ! 2 if f(p) = x. By a similar analysis, G(f) $
m ! 2 if f(p) = y or f(p) = z.
Now suppose that, with p as given in the table above, f(p) = w 0 X\{x, y, z}. Let pO be the
following profile for which pO(2) = p(2) and pO(3) = p(3):
1
2
3
w
y
z
!
z
x
z
x
y
y
!
!
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Then z is a strong Condorcet winner at pO. Therefore, f(pO) = z. But z ranks second last in pO(1),
and at profile pO person 1 can precipitate the selection of her top ranked alternative w by reporting
p(1). Therefore, G(f) $ m ! 2 in this case as well.
Part 2. Assume now that n > 3 and n is odd. Construct p by letting p(1), p(2), and p(3) be
as in Part 1 above. Let Q be any strong ordering on X and set p(i) = Q for any even i > 2. Let R
be the inverse of Q obtained by turning Q upside down. Set p(i) = R for every odd i > 3. Note
that we can partition the individuals in {4, 5, . . . , n} so that half have Q at p and half have R.
Therefore, in determining a strong Condorcet winner (or the absence of one) the individuals i > 3
can be ignored because they offset each other in terms of the vote count. Then the argument that
we employed above for n = 3 can be applied without modification to establish G(f) $ m ! 2.
G
This First Condorcet Theorem allows G(f) = m – 1 or m – 2. Both possibilities can be
realized.
Example 5.1. Let f be the rule that selects the Condorcet winner if there is one and
otherwise picks individual #1's bottom ranked alternative. It can be shown that G(f) = m – 1 (by
manipulations from one non-Condorcet winner to another).
Example 5.2. Let f be the rule that selects the Condorcet winner if there is one and
otherwise picks the alphabetically earliest of the tops. To see that G(f) = m – 2, suppose that á is
at the bottom of u(i) and is chosen. If á is a Condorcet winner, there is nothing i can do to prevent
á from being a Condorcet winner. If there is no Condorcet winner and á is the alphabetically
earliest top, let â be the top of u(i), so á precedes â alphabetically. Then there is nothing i can do
to make â a Condorcet winner and i can not alter the fact that á beats â alphabetically.
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The First Condorcet Theorem does not extend to n = 2 even if we add anonymity, as we
show by the next example. When n = 2, Condorcet is very weak and does not even imply ñ(f) = m
as it does for all larger n. But ñ(f) = m for the next example, which also satisfies a Pareto
condition and anonymity. The example has n = 2, m = 4, and G(f) = 1, although the rule is
Condorcet.
Example 5.3. Let n = 2 and X = {x,y,z,w} with f(u) determined by a sequence of steps:
Step 1. If at profile u, anyone has w at her top, f(u) = w.
Step 2. If at profile u, both individuals have the same alternative á 0 {x,y,z} at their top,
then f(u) = á.
Step 3. If at profile u neither Steps 1 nor 2 apply, consider the following list of orderings:
zw; yw; xw; zy; zx; yz; yx.
At least one individual must have one of those orderings as their initial segment. Pick the
individual with the initial segment of their order, say áâ, earliest in the list. Then f(u) is that
alternative in {á,â} that is preferred by the other individual.
First ñ(f) = 4 as w will be chosen when it is at one person’s bottom and at the other’s top.
It can be shown that only w can be chosen from someone’s bottom and that no one ever
manipulates to w.
It is clear that this rule satisfies anonymity. We next show it satisfies Pareto and hence is
Condorcet. That is, we want to show that if f(u) = á, there is no â such that both prefer â to á.
This is obvious if f(u) = á via Steps 1 or 2. So suppose that f(u) = á by Step 3. Then one
individual has initial segment either áâ or âá. Clearly if the initial segment is áâ, there is no
alternative both prefer to á. But if the initial segment is âá, the only possible alternative both
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prefer to á is â. But if the other individual also prefers â to á, then á wouldn’t have been chosen.
Finally we show G(f) = 1. So suppose that f(u) = á; we want to show no one can improve
to an outcome two or more ranks higher in their ordering.
1. f(u) = á arises from Step 1. An individual with w on their top can not improve at all,
and the other individual can not affect the outcome.
2. f(u) = á arises from Step 2. Each individual has á as their top and so can not improve at
all.
3. f(u) = á arises from Step 3. Some individual has áâ or âá as her initial segment; the
other individual’s initial segment appears later in the list and that individual prefers á to â (and so
can’t gain three ranks). Now the first individual can not improve two ranks (or three) since á is in
their initial segment of length two. The only way the other individual could improve two ranks
would be if they had ordering ãäáâ and could precipitate the selection of ã. But since ãä is later in
the list than the first individual’s initial segment, the other would have to submit an ordering with
one of ãá, ãâ, áã, or âã as initial segment. But then since the first individual prefers each of á and
â to ã, alternative ã wouldn’t be chosen.
The fact that n = 2 in Example 4.3 is crucial for obtaining G(f) = 1.
Open Question: For m $ 3 and even n $ 4, does every Condorcet social choice rule f
satisfy G(f) $ m – 2?
We next show that if we strengthen the Condorcet condition slightly, the answer is “yes.”
With even n, the Condorcet condition is weak. We can strengthen it by requiring that if at profile
u there exist alternatives that are not defeated by majority vote, then f(u) is one of those
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alternatives. An alternative x is a Condorcet non-loser if there is no other alternative y that defeats
x by majority vote (though there may be alternatives that are tied with x). Then, because so many
ties are possible, rules are heavily influenced by how ties are broken. A selection function, t,
chooses an element from each subset Y of X:
t(Y) 0 Y.
A rule f is fully Condorcet if there is a selection function t such that, for each profile u for which
the set Y(u) of Condorcet non-losers is non-empty, f(u) = t(Y(u)). For even n, the selection
function for any fully Condorcet rule is unique. The selection function t need not be “rationalized”
— t(Y) need not be the highest element in some transitive ordering. It is also important to note
here that t(Y(u)) depends only on Y(u), and not separately on u: Hence Y(u) = Y(v) … i implies
f(u) = f(v). For a fully Condorcet rule, Range(f) = X and ñ(f) = m, even if n = 2. (Proof: If t(X) =
x, let Q be a strong order with x at the top. Let R be the inverse of Q. Then consider a profile u
where half of the individuals are assigned Q and half are assigned R.)
Notice that Example 4.3 is not fully Condorcet as its tie-break can not be represented by a
selection function. To see this, look at two profiles; the first is u:
1
2
x
y
z
w
z
y
x
w
The set of Condorcet non-losers is {x,y,z}. Now f(u) = y since #2 has the ordering with initial
segment, zy, earlier in the list than xy and #1 prefers y to z. If f were fully Condorcet via selection
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function t, we would need t({x,y,z}) = y. But look at the second profile, v:
1
2
x
z
y
w
y
z
x
w
Again the set of Condorcet non-losers is {x,y,z}. Since t({x,y,z}) = y, fully Condorcet would
require f(v) = y. But f(v) = z since #2 has the ordering with initial segment, yz, earlier in the list
than xz and #1 prefers z to y.
Theorem 5.2 (Second Condorcet Gain Theorem). For even n $ 2, and any fully
Condorcet rule f, G(f) $ m–2.
Proof: Let t be the selection rule for f. Without loss of generality, we assume that
t({x,y,z}) = x and t({y,z}) = y.
Case 1. n = 4. Let u be a profile such that:
1
x
y
z
!
2
3
y
!
x
z
4
z
x
y
!
z
y
x
!
Then f(u) = t({x,y,z}) = x. If individual #2 manipulates to u* by reporting the ordering yzx..., then
f(u*) = t({y,z}) = y and #2 gains m – 2 ranks.
Case 2. Even n > 4. Construct u by letting u(1), u(2), u(3), and u(4) be as in the table
above. Let Q be any strong ordering on X and set p(i) = Q for any even i > 4. Let R be the
Page -20-
inverse of Q, and set p(i) = R for every odd i > 4. Then the argument above for n = 4 can be
applied without modification to establish G(f) $ m ! 2.
~
Section 6. Scoring rules.
While the previous section addressed one large class of rules, this one takes up another
large class -- scoring rules, which generalize Borda’s rule. Scoring rules are characterized by a
sequence (n1, n2, . . . , nm–1, nm). For any ordering r of X, r[j] is assign the score nj. Then, at profile
u, each alternative x in X is assigned the sum of its u(i) scores over all i 0 N. f(u) is then the
alternative with the smallest sum. Of course, ties are possible, so a complete description of f
requires a tie-break rule. Here we assume that X is ordered (“alphabetically”) and then f(u) is the
alphabetically earliest of the alternatives with the smallest sum. Often the sequence (n1, n2, . . . ,
nm–1, nm) is assumed to satisfy
n1 # n2 # . . . # nm–1 # nm
with at least one inequality strict. Such scoring rules will be called standard.
Example 6.1. (Borda rule) This rule is characterized by the sequence (1,2, ..., m).
Subsection 6.1. Standard rules with two values.
We start by analyzing standard scoring rules that employ just two score values, which,
without loss of generality, we take to be 0 and 1. Consider f where 1 is assigned to the bottom t2
ranks and 0 is assigned to each individual’s top t1 ranks where t1 + t2 = m and ties are broken
alphabetically. This is a kind of approval voting, but one in which the number of approvals is the
same for every individual and every profile.
Theorem 6.1. Let f be a standard scoring rule with two values:
Page -21-
1. If ñ(f) < m, then G(f) = ñ(f) – 1;
2. If ñ(f) = m, then G(f) = Max{t1–1, t2–1}.
Proof: Standard rules with two scores and nt2 $ m were treated in Example 4.2 of Section
4 where it was shown that ñ(f) = m but G(f) = Max{t 1–1, t 2–1}. If nt 2 < m, some alternative is
assigned 0 by everyone, so no-one’s last is ever chosen; in fact, ñ(f) # t1, as we now show.
Consider a profile u. As noted there exist alternatives to which everyone assigns 0; let a be the
alphabetically earliest: f(u) = a. Alter u to u* by (only) lowering a to everyone’s t1 position. No
scores are changed, so f(u*) = a, and thus ñ(f) = t1.
We now determine G(f). Construct profile u:
1
2
þ
n
b
x
!
a
b
x
!
a
b
x
!
a
b
x
!
a
t1
z
!
z
!
z
!
z
!
t2
where a is in everyone’s t1 position. Then f(u) = a. But if u* is created by #1 interchanging a and
z, then f(u*) = b and #1 gains t1 – 1 = ñ(f) – 1 positions. Since f(u) is always in the top t1 ranks for
all individuals, t1 – 1 is the maximum gain and G(f) = t1 – 1.
~
Subsection 6.2. Standard rules with three values. We next analyze scoring rules that
employ three values, which, without loss of generality, we take to be 0, p, and 1 with 0 < p < 1.
Therefore we assume in this section that m $ 3. Value 1 is assigned to the bottom tier, the bottom
Page -22-
t3 ranks, p is assigned to the middle tier, the next t2 ranks and 0 is assigned to the top tier, the top
t1, where t1 + t2 + t3 = m and ties are broken alphabetically.
Theorem 6.2. Let f be a standard scoring rule with three values:
1. If ñ(f) < m, then G(f) = ñ(f) – 1;
2. If ñ(f) = m, then G(f) = Max{t1 + t2 –1, t2 + t3 –1}.
Proof: There are three cases:
A. n(t2 + t3) < m;
B. nt3 $ m;
C. n(t2 + t3) $ m, but nt3 < m;
and we treat these in turn.
Case A. If n(t2 + t3) < m, some alternative must be assigned 0 by everyone, so ñi(f) # t1
for all i and then ñ(f) # t1. Now consider a profile u, with f(u) = x. As noted, x ranks no lower
than position t1 for all i. Alter u to u* by (only) lowering x to everyone’s t1 position. No scores
are changed, so f(u*) = x, and thus ñ(f) = t1.
We now show that G(f) $ ñ(f) – 1. First we establish t1 > 1. Suppose that t1 = 1. Since n
$ 2, we have that n(t2 + t3) < m implies 2(t2 + t3) < m. But t1 = 1 implies t2 + t3 = m – 1, and so
2(m – 1) < m, implying m < 2, a contradiction. Therefore t1 > 1.
Page -23-
Now that we know t1 > 1, construct profile u:
1
2
b
x
!
a
b
x
!
a
z
!
z
!
v
!
v
!
þ
n
b
x
!
a
t1
þ
z
!
t2
þ
v
!
t3
þ
where a is in everyone’s t1 position. Then f(u) = a. But if u* is created by #1 interchanging a and
z, f(u*) = b and #1 gains t1 – 1 positions. Thus G(f) $ t1 – 1 = ñ(f) – 1.
Case B. If nt3 $ m, then ñ(f) = m. To see this, let u be a profile such that every
alternative is in someone’s last t3 ranks. Then x = f(u) is in the last t3 ranks for some i; construct
u* by lowering x to i’s last rank. Then x = f(u*) = f(u) = u*(i)[m]. By anonymity, ñ(f) = m.
No one can manipulate from their last rank to their first. In fact, no one can manipulate
from their bottom tier to their top tier, because one can’t unilaterally reduce the score in one’s top
tier or increase the score of an alternative in one’s bottom tier. So G(f) can not exceed Max {t 1 +
t2 – 1, t2 + t3 – 1}. We will show G(f) = Max {t1 + t2 – 1, t2 + t3 – 1} by showing that f allows
gains of t1 + t2 – 1 and t2 + t3 – 1.
Part 1. We first show f allows a gain of at least t1 + t2 – 1.
Page -24-
Subcase 1. t1 > 1. Consider profile u:
1
2
þ
n–1
n
b
!
z
a
!
b
þ
a
!
b
a
!
y
t1
!
a
!
þ
þ
!
b
t2
v
!
!
þ
þ
!
t3
where every alternative except a and b appears in someone’s last t3 ranks and v appears there at
least twice (recall nt3 $ m). Then the sum of scores for a and b is p while the sum of scores for any
other alternative is at least 1. Since a and b are tied, the tie break gives f(u) = a. Now if #1
interchanges a and v, then b has the smallest sum of scores and #1 gains t1 + t2 – 1 ranks.
Subcase 2. t1 = 1. Suppose first that n is even. Consider profile u:
1
2
þ
b
b
þ
a
a
!
a
!
a
þ
!
b
!
b
v
!
!
þ
þ
!
n–1
n
where every alternative except a and b appear in someone’s last t3 ranks and v appears there at
least twice. Then the sum of scores for a and b is p(n/2) while the sum of scores for any other
alternative is more than that. Since a and b are tied, the tie break gives f(u) = a. Now if #1
Page -25-
interchanges a and v, then b has the smallest sum of scores and #1 gains t1 + t2 – 1 ranks.
For the case of odd n, note first that if t2 = 1 also, then t1 + t2 – 1 = 1 and we know G(f) $
1 by Gibbard-Satterthwaite. So we may assume that t2 > 1. Consider profile u where all
alternatives except a and b appear at least once in the bottom t3 while z appears at least twice in the
bottom t3:
1
þ
þ
n
a
þ
a
z
b
þ
b
!
b
þ
!
b
!
a
b
!
a
þ
!
a
t2
!
z
!
z
!
!
!
!
v
!
t3
(n–1)/2 (n+1)/2 (n+3)/2
At u, alternatives a and b receive scores of p(n+1)/2; alternative z receives only one 0 and at least
two 1's for a score of at least 2 + (n–3)p > p(n+1)/2; all other alternatives receive no 0's and at
least one 1 for a score of at least 1 + (n–1) p > p(n+1)/2. Thus a and b tie for lowest sum and f(u)
= a. But then if the nth individual interchanges a and v, b is chosen and the nth individual gains t1
+ t2 – 1 ranks.
Part 2. We next show f allows a gain of at least t2 + t3 – 1. If t3 # t1, then this is settled by
our proof in Part 1 that G(f) $ t1 + t2 – 1. So we assume that t3 > t1 and in particular, that t3 > 1
and m > 3.
Case 1. t1 > 1. Consider profile u:
Page -26-
1
2
3
þ
n–1
n
c
!
d
c
!
d
a
!
b
þ
a
!
b
a
!
b
b
!
a
!
!
þ
!
!
!
a
!
b
!
þ
!
!
t1
t2
t3
where every alternative appears at least once in someone’s bottom t3 (recall nt3 $ m). Since t3 > t1,
we can ensure that for each alternative in X\{a,b} there are at least as many appearances in the
bottom t3 as in the top t1. Then f(u) = a, on a tie-break with b. But if u* is constructed by
interchanging b and d in #1's ordering, f(u*) = b.
Case 2. t1 = 1. We first treat what happens when n is even. Consider u:
12
c
3
a
b
a
!
b
!
!
!
b
!
c
!
a
þ
n–2
n–1
n
þ
a
b
b
t1
b
!
a
!
!
a
t2
!
!
!
þ
t3
where a and b alternate as tops for individuals 2, 3, ... , n – 1. Every alternative appears at least
once in the bottom t3. With lowest sum, b = f(u). Then #1 gains t2 + t3 – 1 ranks by interchanging
a and c.
If n is odd, Consider profile u:
Page -27-
12
c
3
a
b
b
!
!
a
!
a
!
b
!
c
þ
n–2
þ
ba
þ
n–1
n
b
t1
a
!
b
!
a
!
t2
!
!
!
t3
where a and b alternate as tops for individuals 2, 3, ... , n. Every alternative appears at least once in
the bottom t3 and a and b only appear once in the bottom t3. Now a and b are tied with lowest sum
and f(u) = a. When #1 interchanges b and c, he gains t2 + t3 – 1 ranks.
Case C. n(t2 + t3) $ m, and nt3 < m,
Since n(t2 + t3) $ m, there is a profile u at which, for every alternative, there is an individual
who has that alternative in their bottom t2 + t3 ranks. In particular, someone has f(u) in their bottom
t2 + t3 ranks. If for any such profile f(u) is in someone’s bottom t3 ranks, that profile could be
modified so f(u) is in someone’s mth rank. Anonymity then implies ñ(f) = m. Otherwise, for some
profile u, f(u) is in someone’s middle t2 ranks and that profile could be modified so f(u) is at rank t1
+ t2. Anonymity then implies ñ(f) = t1 + t2. So we separate this Case into two Subcases.
Page -28-
Subcase C1: ñ(f) = t1 + t2. Again we first treat the case where t1 > 1. Consider profile u:
1
2
3
þ
n–1
n
a
!
c
b
!
d
a
!
b
þ
a
!
b
a
!
b
!
b
!
a
!
þ
!
!
t2
!
d
!
c
!
þ
!
!
t3
t1
where every alternative appears at least once in someone’s bottom t2 + t3 (recall n(t2 + t3) $ m).
Alternatives a and b have sum p while all other alternatives have sum at least p. By the alphabetic
tie-break, f(u) = a. If now #2 interchanges a and c, alternative b wins and #2 gains t1 + t2 – 1 = ñ(f)
– 1. Next suppose that t1 = 1 and n is even. Consider profile u:
1
2
3
þ
n–1
n
a
b
a
þ
a
b
t1
!
b
!
a
!
b
þ
!
b
!
a
t2
!
d
!
c
!
þ
!
!
t3
Alternatives a and b each have sum (n/2)p while all other alternatives have sum at least np. By the
tie break, f(u) = a. If #2 interchanges a and c, then b wins and #2 gains t1 + t2 – 1 = ñ(f) – 1.
Page -29-
If t2 = 1 also, then t1 + t2 – 1 = 1 and we know G(f) $ 1 by Gibbard-Satterthwaite. So we assume
that t2 > 1. Consider the profile
1
2
3
þ
n–1
c
a
b
þ
a
b
!
a
!
b
!
a
þ
!
b
!
a
!
d
!
c
!
e
þ
!
!
n
b
t1
t2
t3
Then a and b have sum p(n+1)/2, while all other alternatives have sum at least np. Hence, f(u) = a.
If #3 interchanges a and e, then b wins and #2 gains t1 + t2 – 1 = ñ(f) – 1.
Subcase C2: ñ(f) = m. We want to show G(f) $ Max{t1 + t2 – 1, t2 + t3 – 1}. The
argument above for the case ñ(f) = t1 + t2 also suffices here to show G(f) $ t1 + t2 – 1. We need
only show G(f) $ t2 + t3 – 1. Of course, if t2 + t3 – 1 # t1 + t2 – 1, i.e., if t3 # t1, we are done. So we
assume that t3 > t1, and, in particular, t3 > 1.
We now claim that for standard scoring rules with three scores and fixed n, m, t1, t2, and t3,
such that n(t2 + t3) $ m, nt3 < m, and t3 > t1, we have (1) ñ(f) = t1 + t2 if p < 1/n and (2) ñ(f) = m if
p $ 1/n. To prove the first claim, suppose that p < 1/n and x is in person i’s bottom t3 ranks. Then i
gives x a score of 1. Since nt3 < m, some alternative receives only scores of 0 and p and so a total
no greater than np. Then having x win certainly requires 1 # np, i.e., p $ 1/n.
So p < 1/n implies ñ(f) # t1 + t2. But since n(t2 + t3) $m, there is a profile u such that every
alternative is in someone’s bottom t2 + t3 ranks. By the previous paragraph, f(u) must be in
Page -30-
someone’s middle t2. Bringing that alternative down to rank t1 + t2 doesn’t change the outcome, so
ñ(f) = t1 + t2.
If p $ 1/n, consider profile u:
1
2
3
þ
n–1
n
a
!
a
!
t1
!
j
i
g
!
j
i
t2
!
h
t3
b
!
c
a
!
a
!
þ
d
!
j
i
e
!
j
i
f
!
j
i
þ
!
a
!
!
þ
!
Here any alternative other than a that appears in the top tier appears only once (possible since nt1 <
m) and also appears once somewhere in the bottom tier (possible since t3 > t1). The sum for a is 1.
The sum for any other alternative appearing in the top tier is at least 1 + (n–2)p. The sum for any
alternative that does not appear in the top tier is at least np. Since np $ 1, alternative a at least ties
for least sum and then, by the tie-break, f(u) = a and ñ(f) = m.
So, for Subcase C2, ñ(f) = m and we have p $ 1/n. We must distinguish between those
cases where 1/n # p < 1/(n–1) and those where 1/(n–1) # p < 1.
Subcase C2A. 1/n # p < 1/(n–1).
Consider profile u:
Page -31-
1
2
3
þ
n–1
n
b
!
c
a
!
a
!
þ
a
!
a
!
d
!
f
!
j
d
þ
!
j
d
g
!
j
d
t2
j
e
!
j
d
!
a
!
b
!
þ
!
h
t3
!
t1
Here any alternative other than a that appears in the top tier appears only once (possible since nt1 <
m) and also appear once somewhere in the bottom tier (possible since t3 > t1). The sum for a is 1.
The sum for any other alternative appearing in the bottom tier is at least 1. Since nt3 < m, at least
one alternative appears only in the middle tier with sum np; let d be the alphabetically earliest such
altenative. Since np > 1, f(u) = a. If #1 interchanges b and d, then d’s sum falls to (n–1)p < 1 and d
is chosen, so #1 gains t2 + t3 – 1 ranks.
Subcase C2B. 1/(n–1) # p < 1.
Suppose first that t1 > 1. And consider profile u:
Page -32-
1
2
3
þ
n–1
n
c
!
d
e
!
f
a
b
!
þ
a
b
!
a
b
!
t1
b
!
j
i
a
!
j
i
f
!
j
i
þ
!
j
i
g
!
j
i
t2
!
a
!
b
!
þ
!
h
t3
!
Here any alternative other than a or b that appears in the top tier appears only once (possible since
nt1 < m) and also appears at least once somewhere in the bottom tier (possible since t3 > t1). The
sum for a and b is 1 + p. The sum for any other alternative appearing in the bottom tier is at least 1
+ (n-2)p. (Note n > 2 since 1 > p $ 1/(n – 1).) All other alternatives appear only in the middle tier
with sum np. Since (n-1)p > 1, alternatives a and b are tied for low score and by the tie-break, f(u)
= a. But if #1 interchanges d and b, the outcome changes to b and #1 has gained t2 + t3 – 1 ranks.
Consider t1 = 1. Suppose t2 > 1. Let k be the smallest integer such that 1/k # p < 1. So
(k–1)p < 1 # kp. Let z = n – (k–1) and hence n–z = k–1.
1
2
3
þ
z
z+1
þ
n–1
n
c
e
f
þ
r
a
þ
a
a
t1
b
!
j
i
b
a
!
i
b
a
!
j
þ
b
a
!
i
b
!
j
i
þ
b
!
j
i
b
!
j
i
t2
!
a
!
c
!
e
þ
!
!
h
t3
þ
f
Page -33-
Here any alternative other than a that appears in the top tier appears only once (possible since nt1 <
m) and also appears at least once somewhere in the bottom tier (possible since t3 > t1). The sum for
a is 1 + (z–1)p. The sum for b is np. Note that 1 + (z–1)p # np because 1 # kp. Thus f(u) = a.
But if #1 interchanges b and c, then b’s sum falls to (n–1)p and (n–1)p < 1 + (z–1)p since (k–1)p <
~
1. Thus b now is chosen and #1 has gained t2 + t3 – 1 ranks.
Subsection 6.3. m values.
Now suppose that n1 < n2 < . . . < nm–1 < nm. We will treat here only the case n $ m (which
is the analog of nts $ m for each s, inasmuch as each alternative can appear in each tier).
Theorem 6.3 (Scoring Rule Gain Theorem). Let f be a standard scoring rule with n $ m
and n1 < n2 < . . . < nm–1 < nm. Then G(f) = m–2.
Proof: Suppose that n is even and greater than 2. First, no one can manipulate from their
bottom tier to their top tier, because one can’t unilaterally reduce the score in one’s top tier or
increase the score of an alternative in one’s bottom tier; so G(f) # m–2. Consider profile u:
1
2
3
4
þ
m
m+1
þ
n–1
c
d
b
a
þ
a
b
þ
b
a
b
a
a
b
þ
b
a
þ
a
b
!
!
!
!
!
!
!
þ
!
!
a
b
c
d
þ
z
c
þ
n
where a and b appear just once in the bottom row and every other alternative appears there at least
once. Alternatives a and b have lowest score so, by the tie-break, f(u) = a. If #1 interchanges b and
Page -34-
c (leaving everything else unchanged), the tie is broken and f(u*) = b so #1 gains m–2 ranks.
If n is odd, consider profile u:
1
2
3
4
5
þ
m–1
m
m+1
þ
n–1
n
a
b
c
a
b
þ
b
a
b
þ
a
b
c
a
b
b
a
þ
a
b
a
þ
b
a
d
c
c
d
c
d
c
!
!
þ
!
!
z
d
þ
!
!
!
!
!
!
b
c
a
c
d
þ
y
where a and b appear just once in the bottom row and every other alternative appears there at least
once.
Alternatives a and b are tied for lowest sum, so f(u) = a. If #3 now interchanges b and c
(leaving everything else unchanged), f(u*) = b and #3 gains m–2 ranks.
~
Section 7. Tops-only and vector rules.
In this section we treat a large class of rules that generalize tops-only rules. We say that f is
a tops-only rule if for every two profiles u and v, (u(1)[1], u(2)[1], ... , u(n)[1]) = (v(1)[1], v(2)[1],
... , v(n)[1]) implies f(u) = f(v). We may not have f(u) 0 {u(1)[t1], u(2)[t2], ... , u(n)[tn]}:
Example 7.1. Fix two alternatives x and a in X. f(u) = u(1)[1] unless u(1)[1] = x, in which
case f(u) = a.
Theorem 7.1 (Tops-only Gain). For any non-dictatorial tops-only rule f with |Range(f)| >
2, G(f) $ m – 2.
Proof: Since f is non-dictatorial and |Range(f)| > 2, Gibbard-Satterthwaite tells us f is
manipulable. Suppose f is manipulable by i at profile u to i-variant profile v. Then f(u) is not u(i)[1]
Page -35-
or i has no incentive to manipulate. If necessary, alter u by taking f(u) to the bottom of u(i). If f(v)
is u(i)[1] or u(i)[2], the manipulation yields a gain of at least m – 2. Otherwise, alter u to put f(v) in
the second rank. Then manipulating to v yields a gain of m – 2.
~
Both m – 1 and m – 2 are possible gain values for non-dictatorial tops-only rules f with
|Range(f)| > 2:
Example 7.2. Let f select the alphabetically earliest alternative that is no-one’s top.
Suppose that at u, #1 has a at the top and b at the bottom while everyone else has c on top. Then
f(u) = b. But if #1 reports a preference with c on top, then b is chosen and #1 gains m – 1 ranks.
Example 7.3. Let f select the alphabetically earliest top. Suppose that at u, #1 has c at the
top, a second, and b at the bottom while everyone else has b on top. Then f(u) = b. But if #1
reports a preference with a on top, then a is chosen and #1 gains m – 2 ranks. Since no-one can
manipulate to their top. G(f) = m – 2.
We can generalize: A rule f is a vector rule if there is a vector (t1, t2, ... , tn) of integers each
belonging to {1,2, ..., m}such that for every two profiles u and v, (u(1)[t1], u(2)[t2], ... , u(n)[tn]) =
(v(1)[t1], v(2)[t2], ... , v(n)[tn]) implies f(u) = f(v). For such a vector rule, we may abuse notation a
little and express f (u) as f (u(1)[t1], u(2)[t2], ÿ , u(n)[tn]) An example of a vector rule that is not
tops-only is inverse plurality, Example 2.3.
In the following theorem, given without proof, there is no restriction on the parity of n.
Theorem 7.2 (Vector Rule Gain Theorem). For m > n, Let f be any non-dictatorial vector
rule with |Range(f)| > 3:
(1) If ñ(f) = m, then G(f) $ m – 2;
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(2) If 1 < ñ(f) < m, then G(f) $ ñ(f) – 1.
We conjecture that the Vector Rule Gain Theorem also holds for m # n.
Section 8. Final remarks.
For the benchmark case ñ(f) < m, all our examples and analyses so far confirm Gain
Conjecture #1: G(f) $ ñ(f) – 1. But for ñ(f) = m, G(f) can be as small as m/2 – 1, as illustrated by
three kinds of rules in Section 4. All our examples confirm Gain Conjecture #2: G(f) $ m/2 – 1.
But for most situations with ñ(f) = m, G(f) is either m – 1 or m – 2. Obviously, we need something
beyond ñ(f) to sort out the variety of outcomes when ñ(f) = m.
For all the classes of rules we have considered and, assuming the Gain Conjectures, for all
non-dictatorial rules with at least three alternatives in the range, small maximal gains can only be
achieved if some one individual has a great deal of power in determining the values of f. More
carefully, if G(f) < m/2 – 1, then ñ(f) < m/2 and some individual i can precipitate the chosen
alternative to be from some initial segment of their ordering, u(i)[1:ñ(f)]. And that can be true for
only one individual as, for any j … i, there is a u with u(j)[1:ñ(f)] disjoint from u(i)[1:ñ(f)].
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